p Test dsend with timings injection
p message size is 4 bytes
p process 1 will finish at 0.5+2*4 (send) + 1+0.1*4 (isend) = 9.9s
p process 2 will finish at 0.5+2*4 (time before first send) + 2*(1+0.5*4) (recv+irecv) + 0.005890 (network time, same as before) = 14.505890s
! output sort
p Test dsend with timings injection
p message size is 4 bytes
p process 1 will finish at 0.5+2*4 (send) + 1+0.1*4 (isend) = 9.9s
p process 2 will finish at 0.5+2*4 (time before first send) + 2*(1+0.5*4) (recv+irecv) + 0.005890 (network time, same as before) = 14.505890s
! output sort